Problem: Assume that $S$ is an outwardly oriented, piecewise-smooth surface with a piecewise-smooth, simple, closed boundary curve $C$ oriented negatively with respect to the orientation of $S$. $ \oint_C (x^2\hat{\imath} + y\sin(z) \hat{\jmath} + 3ze^x \hat{k}) \cdot dr$ Use Stokes' theorem to rewrite the line integral as a surface integral. $ \iint_S ( $ $ \hat{\imath} + $ $\hat{\jmath} + $ $ \hat{k} ) \cdot dS$
Explanation: Assume we have a continuously differentiable three-dimensional vector field $F(x, y, z)$, an oriented piecewise-smooth surface $S$, and a piecewise-smooth, simple, closed boundary curve $C$ oriented positively with respect to $S$. Then Stokes' theorem states that we have the equality below: $ \oint_C F \cdot dr = \iint_S \text{curl}(F) \cdot dS$ If $C$ is negatively oriented, the line integral is equal to the negative of the double integral. [What does any of that mean?] When we use Stokes' theorem to translate from line integrals to surface integrals, we know $F$ and we want to find $\text{curl}(F)$. $\begin{aligned} F(x, y, z) &= x^2\hat{\imath} + y\sin(z) \hat{\jmath} + 3ze^x \hat{k} \\ \\ \text{curl}(F) &= \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \\ x^2 & y\sin(z) & 3ze^x \end{pmatrix} \\ \\ &= (-y\cos(z)) \hat{\imath} \\ \\ &+ (-3ze^x) \hat{\jmath} \end{aligned}$ Now that we know the curl of $F$, we can use it to find a surface integral equivalent to the original line integral. Because $C$ is negatively oriented, we put $-\text{curl}(F)$ inside the surface integral: $ \iint_S \left[ y\cos(z) \hat{\imath} + 3ze^x \hat{\jmath} + 0 \hat{k} \right] \cdot dS$